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The problem of Orthogonality: gives $n$ vectors of dimension $k$ and another set of same, can a pair be found with inner product = $0$?

The problem of max product: likewise two sets each $n$ vectors (I forgot to mention in both cases values are binary). We want to find maximal inner product (one vector from set $A$ and one from $B$).

I want to find a reduction between OV to max. Inner product.

My idea: Use negative values, claim its equivalent with negative value and possitive values.

What I mean, OV reduced to max. Inner prod with negative values by multiplying all by $-1$.

Then if maximum is zero... Orthogonality. Else, no orthogonality.

But i need to argue negative max inner product equivalent to max. Inner product.

Maybe there is another way?

If it does not work, also interested in reduction from sat.

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  • $\begingroup$ What kind of reduction are you looking for? Many to one reductions? Turing reductions? $\endgroup$
    – nir shahar
    Jan 9 at 23:54
  • $\begingroup$ Turing reduction will be good, and if possible, many to one. But even turing will assiste me. $\endgroup$
    – V. Prasad
    Jan 9 at 23:56
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One possibility is to take the tensor squares of the vector: replace each vector $x$ with a new vector $\hat{x}$ given by $\hat{x}_{ij} = x_i x_j$ (the vectors have length $k^2)$. We have $$ \langle \hat{x}, \hat{y} \rangle = \sum_{ij} x_i x_j y_i y_j = \langle x,y \rangle^2. $$ Therefore if you know the minimum inner product, you can solve OV. It remains to negate all vectors in one of the sets.

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  • $\begingroup$ This is interesting. But for minimum inner product, there is no need for that simply leave the vectors as they are and then if minimal ip is 0 they are orthogonal and else they are not. My problem begins when i try reducing to maximum inner product i thought they are equivalent easily, but it required more than just multiplying by $-1$ (if simply multiply by $-1$, we get also positive values when multiplied, also to note). $\endgroup$
    – V. Prasad
    Jan 10 at 7:12
  • $\begingroup$ Use $\langle \hat x, - \hat y \rangle = -\langle x,y \rangle^2$. $\endgroup$ Jan 10 at 7:18
  • $\begingroup$ Simple and brilliant. $\endgroup$
    – John L.
    Jan 10 at 7:35
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    $\begingroup$ I gave you a start. You take it from here. $\endgroup$ Jan 10 at 11:44
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    $\begingroup$ hm but still for this i would not need to increase the dimension. leaving the data as is already is reduction from orthogonal to minimal value. increasing the dimension has didnt change $\endgroup$
    – V. Prasad
    Jan 12 at 20:56

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